Question

A small straight conductor $\text{PQ}$ is lying at right angles to an infinite current carrying conductor $\text{XY}$. If the conductor $\text{PQ}$ is displaced on metallic rails of negligible resistance parallel to the conductor $\text{XY}$ then

• A. $\text{Q}$ will be at higher potential and $\text{P}$ will be at lower potential.
• B. $\text{P}$ will be at higher potential and $\text{Q}$ will be at lower potential.
• C. Both $\text{P}$ and $\text{Q}$ will be at the same potential.
• D. No emf will be induced.

Answer 1

The correct option is B $\text{P}$ will be at higher potential and $\text{Q}$ will be at lower potential.

As per the faraday's Law of electromagnetic induction, an emf is induced in a moving conductor is,

$E=\overrightarrow{l}.(\overrightarrow{v}\times \overrightarrow{B})$


We can see that the direction of the magnetic field is inward.

Here, $B,l$ and $v$ are mutually perpendicular.

$\therefore E=Blv$

Induced positive polarity is towards $(\overrightarrow{v}\times \overrightarrow{B})$

$\therefore P$ has $\text{+ve}$ polarity & $Q$ has $-ve$ polarity.

Hence, option $\left(B\right)$ is correct.